ñòð. 75 |

? ?2

? ln ? ? + c3 ? 2 + c4 , c = 2.

2 4

For 11? (2.3) we did not find solutions. A particular solution of 12? (2.3) is

c2

?

? = 2c ? ,

12 . f = c, g = 0,

2

?

c2

?c1 e?1 ? + c2 e?2 ? ,

? > ?2 (1 + c),

?

?

? 4

?

c2

h = e?? (c1 + c2 ?), = ?2 (1 + c),

?

? 4

?

?

? ?? c2

?e (c cos ?? + c sin ??), < ?2 (1 + c),

1 2

4

312 W.I. Fushchych, W.M. Shtelen, S.L. Slavutsky

1 + (c/2) ±(c2 /4) ? ?2 (1 + c) 1 + (c/2)

?1,2 = , ?= ,

2(1 + ?2 ) 2(1 + ?2 )

?2 (1 + c) ? (c2 /4)

?= .

2(1 + ?2 )

A particular solution of 13? (2.3) is

1

13? . ? = ? (c2 + c2 ). (2.4)

f = c1 , g = c2 , h = 0,

21 2

Consider system 14? (2.3). The last equation of 14? (2.3) immediately gives

(2.5)

f = c/?

(as before, c is an arbitrary constant). Substituting (2.5) into the remaining equations

of 14? (2.3) we get

d2 2c d

4 2 (?g) + 1 ? (2.6)

(?g) = 0

d? ? d?

and

1

? ?

4? h + (? + 4 ? 2c)h + h = 0. (2.7)

2

Equation (2.6) can be easily integrated and the result is

?

c1 c2

xc/2 e?x/4 dx + (2.8)

g(?) = .

? ?

In particular, when c = 0, the general solution of equation (2.6) takes the form

c1 ??/4 c2

(2.9)

g(?) = e +.

? ?

Equation (2.7) is in itself an equation for a degenerate hypergeometric function and it

can be rewritten in standard Whittaker form

4x2 w ? (x2 ? 4kx + 4m2 ? 1)w = 0, (2.10)

?

where w = w(k, m, x); k, m are parameters, by the substitution

c c?

h(?) = ? (c?2)/4 e??/8 w ,? , (2.11)

.

4 44

When c = 0, the substitution

?

h(?) = e?? Z0 (? ),

? (2.12)

?=

8

reduces (2.7) to the modified Bessel equation of null order, that is

? ?

? ? ?

? Z 0 + Z 0 ? ? Z0 = 0. (2.13)

Reduction and exact solutions of the Navier–Stokes equations 313

Summarizing results (2.5)–(2.12) we can write down the general solution of 14? (2.3)

as follows

?

c c1 c2

xc/2 e?x/4 dx +

?

14 . f= , g= ,

? ? ?

(2.4)

?

c2

c c? 1

(c?2)/4 ??/8

,? , ?=? 2

h=? e w , + g (y)dy + c3 .

4 44 2? 2

(We continue to numerate solutions of reduced NS equations 1? –19? (2.3) as n? (2.4),

where n? = 1? –19? indicates the corresponding ansatz of table 1.) When c = 0 we get

from 14? (2.4) the following particular solution of 14? (2.3)

c1 ??/4 c2

+ , h = e??/8 Z0 (?/8),

14?? . f = 0, ?

g=

e

? ?

(2.4)

c2 ? e?y/2 ? ?y/2

c2 e

?=? 2 1

+ dy + c1 c2 dy + c3 ,

y2 y2

2? 2

?

where Z0 is modified Bessel function satisfying equation (2.12).

Consider system 15? (2.3). The last equation in it gives

c 1

?. (2.13)

f=

?2

The rest equations of 15? (2.3) take the form

d2 c d

(?g) + 1 ? (2.14)

2 (?g) = 0,

d? 2 ? d?

2

c 1

+ g2 ? , (2.15)

2? =

?

? 4

1 c 1

? ?

?+1? h ? h = 0. (2.16)

?h +

2 2 8

Equations (2.14), (2.15) can be easily integrated and the result is as follows

?

c1 c2

xc/2 e?x/2 dx + (2.17)

g= ,

? ?

?

c2

1 1

g 2 (y)dy ? ? ?. (2.18)

?=

2 2? 8

Equation (2.16) is reduced to the Whittaker equation (2.10) by the substitution

c?3 c ?

h(?) = ? (c?2)/4 e??/4 w ,? , (2.19)

.

4 42

Note, when c = 3, function w 0, ? 3 , ? is reduced to the modified Bessel function

42

??3/4 (?/4). The general relation is (Bateman and Erdelyi [2])

Z

v

? (2.20)

w(0, m, x) = xZm (x/2).

314 W.I. Fushchych, W.M. Shtelen, S.L. Slavutsky

So, we can write down the general solution of reduced ns equations 15? (2.3) in the

form

?

c 1 c1 c2

xc/2 e?x/2 dx +

?

f= ? ,

15 . g= ,

?2 ? ?

(2.4)

c?3 c ? ?

c2

1 1

(c?2)/4 ??/4

,? , g (y)dy ? ? ?,

2

h=? e w , ?=

4 42 2 2? 8

where w satisfies the Whittaker equation (2.10).

Consider system 16? (2.3). The two last equations of it give rise to

c?

(2.21)

h = c, ?= + c1 .

2

Taking into account (2.21) we can rewrite the rest equations of system 16? (2.3) as

follows

1 1

? ? ? c f? + f = 0, (2.22)

f+

2 2

1 1

? ? c g + g = 0. (2.23)

g+

? ?

2 2

By substituting

1

??c (2.24)

f (?) = F (? ), ?=

2

into (2.22), we obtain the following equation:

d2 F dF

(2.25)

+ 2? + 2F = 0.

2

d? d?

The general solution of (2.25) is

?

?? 2 2

ey dy . (2.26)

F (? ) = e c2 + c3

Summarizing results (2.21)–(2.26) we write down the general solution of equations

16? (2.3):

(?/2)?c

2

? 2

?

f = exp ? ?c ey dy ,

16 . c2 + c3

2

(?/2)?c

2

? (2.4)

2

g = exp ? ?c ey dy ,

c4 + c5

2

c?

h = c, ? = + c1 .

2

Reduction and exact solutions of the Navier–Stokes equations 315

In the same way we find solutions of reduced equations 17? –19? (2.3). The soluti-

ons are as follows

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